Страница 193 номер 634, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(2\cos^{2}{2x} + 3\sin{4x} + 4\sin^{2}{2x} = 0\)
\(2\cos^{2}{2x} + 6\sin{2x}\cos{2x} + 4\sin^{2}{2x} = 0,\)
\(2 + 6\operatorname{tg} {2x} + 4\operatorname{tg}^{2}{2x} = 0,\)
\(\operatorname{tg} {2x} = z, 4z^{2} + 6z + 2 = 0,\)
\(z_{1{,}2} = \dfrac{-6 \pm \sqrt{36 - 32}}{8},\)
\(z_{1} = - \dfrac{1}{2}, z_{2} = - 1\)
\(\operatorname{tg} {2x} = - \frac{1}{2}\) или \(\operatorname{tg} {2x} = - 1\)
\(x_{1} = - \dfrac{1}{2}\operatorname{arctg} \dfrac{1}{2} + \dfrac{\pi}{2}n,\)
\(x_{2} = - \dfrac{\pi}{8} + \dfrac{\pi}{2}n, n \in Z;\)

2) \(1 - \sin x\cos x + 2\cos^{2}x = 0\)
\(\sin^{2}x - \sin x\cos x + 3\cos^{2}x = 0,\)
\(\operatorname{tg}^{2}x - \operatorname{tg} x + 3 = 0,\)
\(\operatorname{tg} x = z, z^{2} - z + 3 = 0,\)
\(D < 0\) – корней нет;

3) \(2\sin^{2}x + \frac{1}{4}\cos^{3}{2x} = 1\)
\(2\sin^{2}x - \sin^{2}x - \cos^{2}x + \dfrac{1}{4}\cos^{3}{2x} = 0,\)
\(- \cos{2x} + \dfrac{1}{4}\cos^{3}{2x} = 0,\)
\(\cos{2x}\left( \dfrac{1}{4}\cos^{2}{2x} - 1 \right) = 0,\)
\(\cos{2x} = 0\) или \(\cos^{2}{2x} = 4\) – корней нет;
\(x = \dfrac{\pi}{4} + \dfrac{\pi}{2}n, n \in Z;\)

4) \(\sin^{2}{2x} + \cos^{2}{3x} = 1 + 4\sin x\)
\(\sin^{2}{2x} - \sin^{2}{3x} = 4\sin x,\)
\(\left( \sin{2x} - \sin{3x} \right)(\sin{2x} + \sin{3x}) = 4\sin x,\)
\(2\sin\left(-\dfrac{x}{2} \right)\cos\dfrac{5x}{2} \cdot 2\sin\dfrac{5x}{2}\cos\left(-\dfrac{x}{2} \right) = 8\sin\dfrac{x}{2}\cos\dfrac{x}{2},\)
\(2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\left( 4 + 2\sin\dfrac{5x}{2}\cos\dfrac{5x}{2} \right) = 0,\)
\(\sin x = 0\) или \(4 + \sin{5x} = 0\) – корней нет;
\(x = \pi n, n \in Z.\)